1
151 WebCalc Fall 2002-copyright Joe Kahlig
In Class Questions
MATH 151-Fall 02
October 17
What WebCalc calls The Differential Method is actually referred to by Stewart as
linear approximation or tangent line approximation of f at a.
f (x) ≈ f (a) + f 0 (a)(x − a)
If the formula is rewritten as L(x) = f (a) + f 0 (a)(x − a) then this called linearization of f at a.
Now Stewart’s definition of differentials is:
Let y = f (x), where f is a differentiable function. Then the differential dx is an independent variable: that is, dx can be given the value of any real number. The differential
dy is then defined in terms of dx by the equation dy = f 0 (x)dx
dy
This formula is found by starting with dx
= f 0 (x) and multiplying by dx.
The picture shows the differentials and relates them to ∆y and ∆x.
Now f (a + ∆x) = f (a) + ∆y. This can be
seen from the picture. Also note that as
∆x → 0 then dy gets closer to ∆y. Hence
f (a) + dy
f (a + ∆x)
∆y
f (a)
f (a + ∆x) = f (a) + ∆y ≈ f (a) + dy
dx = ∆x
This new approximation is not different from
the tangent line approximation. (just takes a
little bit of algebra and realizing that x − a is
just dx).
a
a + ∆x
We can extend the approximations from lines to other polynomials.
The quadratic approximation to f (x) neat a is
f 00 (a)
f (x) ≈ f (a) + f 0 (a)(x − a) +
(x − a)2
2!
The cubic approximation to f (x) neat a is
f 00 (a)
f 000 (a)
f (x) ≈ f (a) + f 0 (a)(x − a) +
(x − a)2 +
(x − a)3
2!
3!
The approximations are called Taylor Polynomials about x = a. (You learn this in Cal II.)
1. Use differentials to approximate the following.
(a) cos(59o ). Don’t forget to do everything in radians.
y = cos(x) and dy = − sin(x)dx.
Using a = π3 and ∆x = dx = −1o =
√
−π
dy = − sin( π3 ) ∗ 180
= π3603
cos (59o ) = cos
π
3
+
−1π
180
π
3
≈ cos
+
√
π 3
360
=
1
2
+
(c)
π
3
+
π
180
≈ sin
π
3
+
π
360
=
√
3
2
+
gives
√
π 3
360
(b) sin(61o ). Don’t forget to do everything in radians.
y = sin(x) and
dy = cos(x)dx. Using a = π3 and ∆x = dx = 1o =
π
π
π
dy = cos 3 ∗ 180
= 360
sin(61o ) = sin
−π
180
π
180
gives
π
360
√
4
16.1
√
y = 4 x and dy = 4x13/4 dx. Using a = 16 and dx = ∆x = .1 gives dy =
√
√
4
1
1
16.1 ≈ 4 16 + 320
= 2 + 320
1
320
dy
2
151 WebCalc Fall 2002-copyright Joe Kahlig
2. Find the linear approximation of f (x) =
√
√
x + 7 at a = 2 Use it to evaluate 9.06
The equation of the tangent line for f (x) at a = 2 is
y − f (2) = f 0 (2)(x − 2)
y − 3 = 16 (x − 2)
y = 83 + x6
so the linear approximation is L(x) =
Since
8
3
+
x
6
√
√
√
9.06 = 2.06 + 7, we need to compute L(2.06) to find the approximation of 9.06.
L(2.06) =
8
3
+
2.06
6
3. Use Newton’s Method to find the solutions to x5 = 5x − 2 accurate to 4 decimal places.
f (x) = x5 − 5x + 2 and f 0 (x) = 5x4 − 5.
An+1 = An −
a5n −5An +2
5A4n −5
There are three solutions for this formula.
x = 1.3718
x = 0.4021
x = −1.5820